Calculus and Questions - Assignment Example

You must show your work to receive full credit. If you do not show your work, you may earn only partial or no credit at all f(x) = 0 Find a function  such that it is both even and odd. You must prove your choice is correct in order to receive full credit. f (2) = f(-2) = 0 The above argument shows that the function is even Similarly f (-2) = -f(2) = 0 The above argument shows that the function is odd Hence the function is both even and odd. The function is both even and off. 2) C Find the slope of  in the figure below if  is the midpoint of :  Coordinate of M = (x/2,y/2) Find the slope of OM which gives us y/x hence the correct answer is option C. 3) C If  and , then when is ? You must show your work in order to receive credit.      f(g(x)) = a(cx + d) + b acx + ad +b g(f(x)) = c(ax + b) + d acx + bc +d When , bc + d = ad + b This is only true when f(d) = g(b) f(d) = ad + b g(b) = bc + d Hence Option C is the correct answer. 4) B Two scientists positioned at  and  are 3 miles apart simultaneously measure the angle of elevation of a hot air balloon to be  and , respectively. Suppose the balloon is directly above a point on the line segment connecting  and . Find the elevation of the balloon.      We can draw the triangle that fits this situation. Then we can form simultaneous equations using the sin rule: (3-x)/sin 40 = h / sin 50 x / sin 30 = h / sin 60 by manipulating the second equation we get x = 0.577h Substitute this in equation 1 and we get 3 – 0.577h = 0.84y 3 = 1.417h h = 3/1.417 = 2.1 miles hence the correct answer is option B. 5) A True/False. If  is continuous at , then  and  are both continuous at . If true, then explain why; if false, give a counterexample.     This comes from the properties of limits. The limit of the product of two functions is the product of the limits. Since in this case both the functions have a limit at x=0 and are continuous, their product is also continuous. 6) A True/False. If  and  are both continuous at , then  is continuous at . If true, then explain why; if false, give a counterexample.     This is true. This is because the composition of a function does nothing but replace the value of x by another function and since both the functions are continuous at x=0, the resulting composite function is also continuous. 7) A True/False. Suppose a function  is continuous and never equal to 0 over an interval , then the sign of  never changes on that interval .     This is true because since the function is continuous, it will have to cross zero in order to change sign. 8) D Suppose Find all the points where  is continuous. (You must correctly explain your answer in order to receive any credit.) The function is nowhere continuous. Because between any two whole numbers, is a stream of rational and irrational numbers. Therefore if we use the epsilon-delta definition of the limit, the left hand and the right hand limits would be different depending on the rational and irrational values. 9) B Find the trigonometric limit: When 0 is directly plugged into the expression, we get an indeterminate form of 0/0. Apply L’Hospital’s rule and differentiate both numerator and denominator. Plug in zero and we get (5/4). Therefore option B is the correct answer. 10) D Tell where the following function is continuous on the interval  (Give your answer in interval form.) 11) C If  and , then find . a + b = 2 a – b = -2 where a is the limit of f(x) and b is the limit if g(x) add both equations 2a = 0 a=0 hence the limit of the product of the functions is also zero. 12) B Simplify the following expression, if  Since , the term a-b would result in a negative number, and because there is a modulus sign there, we can rewrite it as (a + b) / 2 + (b – a) / 2 which gives us the answer as b. Hence the correct answer is option B. 13) B Find the limit , if  and  in an open interval  containing the point  except possibly  itself, then Directly plugging in the value of 0 would result in the indeterminate form of 0/0. Hence we have to apply L’hospital’s rule, which gives us: cos(g(x)) * (d/dx(g(x)) / (d/dx(g(x)) The differentials cancel resulting in the expression: cos(g(x)). Since g(x) when x = 0 is zero the required limit L is cos(0) =1. Hence the correct answer is option B. 14) C Compute the limit Directly plugging in x=1 in the expression results in the indeterminate form of 0/0. Therefore apply L’Hospital’s rule which will give us the expression: [-0.5cos(1-sqrt(x))][1/sqrt(x)] Plugging in the value of 1 gives us -1/2. Hence the correct option is C. 15) B State whether the function  attains a maximum or minimum value (or both) on the interval . dy / dx = x^3 – 3x^2 + 2x This is zero when x = 0, 1 or 2. Hence the minimum or the maximum lies on these values. Since 2 is not included in the interval the only values to check are 0 and 1. d^2y/dx^2 = 3x^2 – 6x +2 when x = 0, the double derivative is positive, indicating a minimum. When x = 1, the double derivative is negative, indicating a maximum. Hence the correct option is B. 16) C Suppose . Find the equation of the tangent line at the point (0, -1). Find the gradient at the given point. Using the quotient rule, we find the derivative. Since the numerator turns out to be zero the gradient of the tangent line at the given point is zero. Now, by using the intercept form of the straight line, we find the equation to be y = -1. Hence option C is the right answer. 17) B Find the point(s) on the curve , where the slope is . Differentiate the equation to find out the tangent. The equation of the tangent when equated to 5.5 is 1 + 2x^2 = 10x^2 This gives us x^2 = 1/9 X = 1/3 or -1/3 Plug both of these values in the original equation to get the coordinate pair. The correct answer is B. 18) E Find the rate of change in the angle of elevation (in radians/sec) of the camera shown in the Figure 2 at 10 seconds after lift-off. Figure 2 Rate of change in the angle of elevation = d/dt * ds/dt We can easily find d/dt since = arctan(s/2000) There is incomplete information in the question. In order to solve this question, it is necessary to have the equation of motion of the rocket. 19) B In Figure 3, a 7  connection rod is fastened to a crank of radius 3 . The crank shaft rotates counterclockwise at a constant rate of 200 revolutions per minute. Find an approximation to the velocity of the piston when . Figure 3: The velocity of the crankshaft is related to the angle of the crankshaft 20) Suppose that a function  satisfies the following conditions for all real values of  and : Show that . Use the product rule on the second equation to find f’(x) f’(x) = g(x) + xg’(x) now take the example of a case where x approaches zero then f’(x) = 1 + x similarly f(x) = 1+x hence proved that f’(x) = f(x) Read More